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Evaluate the limit: $$ \lim_{x\to 0}\frac{x-\tan x}{x^3} $$

I solved it like this,

$$ \lim_{x\to 0} \left({1\over x^2} - \frac{\tan x}{x^3}\right) =\lim_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right) $$

Now using the property $$ \lim_{x\to 0} \frac{\tan x}{x}=1 $$

we have:

$$ \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{x^2}\right)=0 $$

Please explain my error! How can I avoid such errors? I have it's correct solution. All I want to know is what I did wrong here?

Note: English is my second language.

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    $\begingroup$ I'm not going to comment on every answer, but all answers so far except mine make no effort at answering the actual question: "I have it's correct solution.All I want to know is what I did wrong here?". $\endgroup$ – Git Gud Feb 13 at 11:48
  • $\begingroup$ $1 = x (1/x)$. Now using the property $\lim_{x \to \infty} 1/x = 0$ we have $1 = \lim_{x \to \infty} x (1/x) = \lim _{x \to \infty} x \cdot 0 = \lim_{x \to \infty} 0 = 0$. $\endgroup$ – Matthew Towers Feb 13 at 15:53
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    $\begingroup$ In general you can't replace a part of the expression by its limit unless the part occurs as a term or a factor in whole expression. Moreover if the part occurs as a factor then it's limit should be non-zero before we can replace it by its limit. In your case $(\tan x) /x$ is neither a factor nor a term in the expression $$\dfrac{1}{x^2}-\dfrac{\tan x} {x} \cdot\dfrac{1}{x^2}$$ See more details at math.stackexchange.com/a/1783818/72031 $\endgroup$ – Paramanand Singh Feb 15 at 2:40
  • $\begingroup$ @ParamanandSingh In the second type "You can replace sub-expression S by its limit L if L≠0...." Why do we need to check that L=/= 0? $\endgroup$ – Navneet Kumar Feb 15 at 8:05
  • $\begingroup$ When we try to prove these rules the restriction $L\neq 0$ is crucial for second case. See proof with more theoretical details at math.stackexchange.com/q/2971122/72031 $\endgroup$ – Paramanand Singh Feb 15 at 8:07
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The explanation you're looking for is this. You are implicitly using the properties $\lim\limits (f+g)=\lim\limits(f)+\lim\limits(g)$ and $\lim\limits (fg)=\lim\limits(f)\lim\limits(g)$, but this is only true when all the limits in these equalities exist, (disclaimer: there are important assumptions that I'm not writing but that should accompany these properties). More specifically, what you did (implicitly) was: $$ \begin{align} \lim\limits_{x\to 0} \left({1\over x^2} - \frac{\tan x}{x^3}\right) &=\lim\limits_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right)\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) + \lim\limits_{x\to 0}\left(- {\tan x\over x}\cdot {1\over x^2}\right) \tag{Incorrect}\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) - \lim\limits_{x\to 0}\left( {\tan x\over x}\right)\lim\limits_{x\to 0}\left({1\over x^2}\right) \tag{Incorrect}\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) - \lim\limits_{x\to 0}\left({1\over x^2}\right) \tag{*}\\ &=\lim\limits_{x\to 0}\left({1\over x^2} - {1\over x^2}\right) \tag{Incorrect}\\ &=0 \tag{**} \end{align} $$

$(\text*)\text{ As correct as something meaningless can be}$
$(\text{**})\text{ Actually correct, but it's too late}$

You can't just replace the value of the limit inside without using the above reasoning or something else which will end up not working.

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    $\begingroup$ Very nice step by step presentation indicating the (in)validity of each step. +1 $\endgroup$ – Paramanand Singh Feb 15 at 2:38
  • $\begingroup$ Hahaha. Comment, down voter? $\endgroup$ – Git Gud Feb 26 at 21:45
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Although, in fact, $\lim_{x\to0}\frac{\tan x}x=1$, you cannot deduce from that that$$\lim_{x\to0}\frac1{x^2}-\frac{\tan x}x\times\frac1{x^2}=\lim_{x\to0}\frac1{x_2}-\frac1{x^2}.$$

In this case, L'Hopital's Rule is the way to go:\begin{align}\lim_{x\to0}\frac{x-\tan x}{x^3}&=\lim_{x\to0}\frac{-\tan^2x}{3x^2}\\&=-\frac13\left(\lim_{x\to0}\frac{\tan x}x\right)^2\\&=-\frac13.\end{align}

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  • $\begingroup$ But,I want to know why can't I directly put tanx/x = 1 while evacuating the limit? $\endgroup$ – Navneet Kumar Feb 13 at 11:23
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    $\begingroup$ Because, as you proved it, it leads you to false statements. $\endgroup$ – José Carlos Santos Feb 13 at 11:23
  • $\begingroup$ Forgive me, but I cannot approve this answer. You may not write $\lim$ before you have established the existence of the limit. Strictly speaking what you have written is a circular argument. There is an easy way out. Write the fraction as $T(x)/N(x)$. Recognize that $T(x)=N(x)=0$. Investigate $T'(x)$ and $N'(x)$. Discover that $T'(x)/N'(x)$ has a limit. Invoke l'Hospital's rule. Conclude that $T(x)/N(x)$ has a limit and that the two limits are identical. This is longer, but clean and cannot be faulted. $\endgroup$ – Carl Christian Feb 13 at 21:59
  • $\begingroup$ I disagree. L'Hopital's Rule states that if $\lim_{x\to0}f(x)=\lim_{x\to0}g(x)=0$ and if $\lim_{x\to0}\frac{f'(x)}{g'(x)}$ exists, then the limit $\lim_{x\to0}\frac{f(x)}{g(x)}$ exists too and it is equal to the previous one. Therefore, the fact that, in the end, I found a limit justifies that the limit $\lim_{x\to0}\frac{f(x)}{g(x)}$ exists. So, when working in the context of l'Hopital's Rule, there is no need to establish from the start the existence of the limit. But it was nice of you to have told me the reason of your downvote. $\endgroup$ – José Carlos Santos Feb 13 at 23:51
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    $\begingroup$ From a rigor point of view, I agree with José. Equality is symmetric, it doesn't matter in what order we write equalities. From a pedagogy point of view, I think @CarlChristian makes a very good point. $\endgroup$ – Git Gud Feb 15 at 18:09
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Right, $$\lim_{x\to0}\frac{\tan x}x=1.$$

But that doesn't mean that you can replace $\dfrac{\tan x}x$ by $1$ inside the limit !

Actually,

$$\frac{\tan x}x=1+f(x)\ne1$$ and the function $f$ can strike back.


The "striking back" works like this:

  • subtracting $1$ from $\dfrac{\tan x}x$ isolates $f(x)$.

  • then dividing by $x^2$ "amplifies" it, giving the term $\dfrac{f(x)}{x^2}$. It turns out that $f(x)$ has a double root at $x=0$, so that the fraction bring a finite contribution, but it could very well have been unbounded.

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Everyone here is correctly pointing out that you cannot use $\displaystyle \lim_{x\to 0} \dfrac{\tan x}{x}=1$, but they haven't stated why is that.

Now let me try to explain it to you.

Before proceeding I would like to clear a fact.

Let $f(x)$ and $g(x)$ be two functions such that $\displaystyle \lim_{x\to a}f(x)=l$ and $\displaystyle \lim_{x\to a} g(x)=m$ then,

$$\displaystyle \lim_{x\to a}\left(f(x)\pm g(x)\right) =l\pm m $$

But the converse i.e.

$$l\pm m = \displaystyle \lim_{x\to a}\left(f(x)\pm g(x)\right)\ \ \ ....(1)$$

may not be true.

I have marked this as $(1)$ because I'll be referring to it later.

You did this

$$ \lim_{x\to 0} \left({1\over x^2} - \frac{\tan x}{x^3}\right) =\lim_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right) $$

Everything looks fine uptill here. Now you want to use this property $\displaystyle \lim_{x\to 0} \dfrac{\tan x}{x}=1$

So you will have to split this expression like this.

$$ \lim_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right) $$

$$\implies \lim_{x\to 0}{1\over x^2} - \lim_{x\to 0}\left({\tan x\over x}\cdot {1\over x^2}\right) $$

$$\implies \lim_{x\to 0}{1\over x^2} - \lim_{x\to 0}{\tan x\over x}\cdot\lim_{x\to 0} {1\over x^2} $$

Now you can use $\displaystyle \lim_{x\to 0} \dfrac{\tan x}{x}=1$ to get this

$$ \lim_{x \to 0}\frac{1}{x^2} - \lim_{x \to 0}\frac{1}{x^2}$$

Upon reaching this stage, you did this step

$$ \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{x^2}\right) $$

But this is wrong as I already stated in point (1) above.

Now you'd why is that. It is because

$$\lim_{x\to 0}{\tan x\over x}\cdot\lim_{x\to 0} {1\over x^2} \neq \lim_{x \to 0} \frac{1}{x^2} $$

but

$$\lim_{x\to 0}{\tan x\over x}\cdot\lim_{x\to 0} {1\over x^2} = 1.00000000000..........0001 \cdot \lim_{x \to 0} \frac{1}{x^2} $$

This number is so close to $1$ that we just approximate it to $1$.

Therefore

$$\lim_{x\to 0}{1\over x^2} - \lim_{x\to 0}{\tan x\over x}\cdot\lim_{x\to 0} {1\over x^2} = \lim_{x\to 0} {1\over x^2} - 1.00000000000..........0001 \lim_{x \to 0} \dfrac{1}{x^2} $$

That's why I said in point $(1)$ the converse may not be true.

If you properly evaluate the limit as @Jose Carlos Santos and @roman did. You will find the answer to be $-\dfrac{1}{3}$.

You can clearly note that this answer is negative, because $$1.00000000000..........0001 \lim_{x \to 0} \frac{1}{x^2} > \lim_{x\to 0} {1\over x^2} $$

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It has been shown in other answers what exactly went wrong with your approach.

As an alternative consider the Taylor expansion of your function under the limit around $0$. It's known that: $$ \tan x \sim x + {x^3 \over 3} + O(x^5) $$

Note that: $$ \frac{\tan x}{x} \sim {1\over x}\left(x + {x^3 \over 3} + O(x^5)\right) = 1 + {x^2\over 3} + O(x^4) $$

Thus as noted by Yves Daoust: ${\tan x \over x} = 1 + f(x)$.

Using this your limit becomes: $$ \lim_{x\to0} \frac{x - \tan x}{x^3} \sim \lim_{x\to0}\frac{x - x - {x^3\over 3}}{x^3} = -{1\over 3} $$

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  • $\begingroup$ Those should be big O’s, not little o’s. $\endgroup$ – Emil Jeřábek Feb 13 at 14:22
  • $\begingroup$ @EmilJeřábek you are right, thanks. $\endgroup$ – roman Feb 13 at 14:35
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Another way of solving the problem is by using series expansion at $x=0$, then you have $\lim_{x \rightarrow 0} \frac{x- \tan{x}}{x^3} = \lim_{x \rightarrow 0} \frac{x-(x + \frac{x^3}{3} + O(x^5))}{x^3} = \lim_{x \rightarrow 0} \frac{-\frac{x^3}{3}+O(x^5)}{x^3} = -\frac{1}{3}$

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Note that you can only split limits involving two functions if limits of both functions exists individually (and must be a finite, well defined numbers).

Therefore you first step itself is not allowed while evaluating limits. Moreover you can't replace (tanx)/x by 1 inside limit, you will have to first split the limit, but as I have mentioned it's an incorrect operation and would yield absurd results if ignored.

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